# Physics Important Questions: Kinetic Theory of Gases

 Maharashtra State Board > Science >  Kinetic Theory of Gases > You are Here

### Very Short Answers (1 Mark)

Q1. Which type of ideal gas will have the largest value for [CP-CV]?

• Cp – Cv will be same for all ideal gas. Because for ideal gas by Mayer’s relation Cp – Cv = R where R is the universal gas constant.

Explanation (not required in exam):

• The specific heat at constant volume (Cv) for monoatomic gas is, Cv = 3/2 R
• The specific heat at constant volume (Cv) for diatomic gas is, Cv = 5/2 R
• The specific heat at constant volume (Cv) for polyatomic gas is, Cv = 3 R

But we know that, for an ideal gas, Cp – Cv = R

So,the specific heat at constant pressure (Cp) for monoatomic gas will be,

Cp = Cv + R = 3/2 R + R = 5/2 R

∴  Cp – Cv for monoatomic gas will be, Cp – Cv = 5/2 R – 3/2 R

the specific heat at constant pressure (Cp) for diatomic gas will be,

Cp = Cv + = 5/2 R + R = 7/2 R

∴ Cp – Cv for monoatomic gas will be, Cp – Cv = 5/2 R – 3/2 R

the specific heat at constant pressure (Cp) for polyatomic gas will be,

Cp = Cv + = 3 R + R = 4 R

∴  Cp – Cv fo polyatomic gas will be, Cp – Cv = 4 R – 3 R

Thus the value of Cp – Cv will be same for all ideal gas.

Q2.  State the equation of vibrational energy of a diatomic molecule.

• Non-rigid molecules of diatomic gas have vibrational energy. It is given by

Evibr = KBT

Where T is the absolute temperature of the gas and KB is Boltzmann’s constant.

Q3. What are the value of emissivity of perfectly (a) black body and (b) reflecting body?

• The emissivity is denoted by letter ‘e’ and it is unitless quantity.  For perfectly black body e = 1, for perfect reflector e = 0 and for all other bodies 0 < e < 1.

Q4. A good absorber is also a good emitter of heat. Correct or incorrect? Why`?

• According to Kirchoff’s law of radiation, the statement is correct. Kirchoff’s law of radiation states that “At any given temperature, the emissivity (or coefficient of emission) of a body is equal to the coefficient of absorption”.
• Thus a = e. Which means a good absorber is also a good emitter of heat.

Q5. The top of a cloud of smoke holds together for hours. Why?

Q6. What is the specific heat of gas in the isothermal process?

• The specific heat of a gas is defined as the amount of heat required to raise the temperature of a unit quantity (1 mole or 1 kg) of gas through 1 K (or 1°C).
• For the isothermal process, there is no change in the temperature. Hence there is no meaning to the phrase the specific heat of gas in the isothermal process

Q7. What is the specific heat of a substance at its boiling point or melting point?

• At constant pressure, the boiling of liquid and melting of solid are isothermal processes.
• The specific heat of a gas is defined as the amount of heat required to raise the temperature of a unit quantity (1 mole or 1 kg) of gas through 1 K (or 1°C).
• For the isothermal process, there is no change in the temperature. Hence there is no meaning to the phrase the specific heat of a substance at its boiling point or melting point.

Q8. What is the origin of pressure exerted by the gases on the wall of the container?

• The molecules of gas are in continuous random motion. During random motion, they continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert a small force on the wall
• The pressure exerted by the gas is due to the sum of all these forces due to the collision of molecules with the walls of the container. The more particles that hit the walls, the higher the pressure.

Q9. What is the average velocity of molecules of an ideal gas?

• The molecules of gas are in continuous random motion. I.e. they are moving in all possible direction in all possible velocities.
• Thus by probability theory the vector sum of velocities of all molecules is zero. Hence the average velocity of the molecules of an ideal gas is zero.

### S.A. – I (2 Marks)

Q1.  State the types of degrees of freedom of rigid diatomic molecules.

• The molecules of rigid diatomic gases have translational motion ( three translational degrees of freedom). The molecule can move along the x-axis or the y-axis or the z-axis.

• Besides translational motion, the molecule has rotational motion(two rotational degrees of freedom). he diatomic molecule can rotate about any axis at right angles to its own axis. Hence it has two degrees of freedom of rotational motion.

• Thus there are 5 degrees of freedom for rigid diatomic molecules.

Q2.  State the type of degrees of freedom of non-rigid diatomic molecules.

• The molecules of non-rigid diatomic gases have translational motion ( three translational degrees of freedom). The molecule can move along the x-axis or the y-axis or the z-axis.

• Besides translational motion, the molecule has rotational motion(two rotational degrees of freedom). he diatomic molecule can rotate about any axis at right angles to its own axis. Hence it has two degrees of freedom of rotational motion.

• Besides translational motion and rotational motion, the molecules have vibrational motion. The molecules can vibrate along the axis of the molecule.
• Thus there are 6 degrees of freedom for non-rigid diatomic molecules.

Q3. Show that for a diatomic gas, the total energy possessed by it is U =(7/2)RT.

• In equilibrium, the total energy is equally distributed in all possible energy modes, with vibrational mode has KBT of energy and each other mode having an average energy equal to

• The molecules of non-rigid diatomic gases have translational motion ( three translational degrees of freedom) as well as rotational motion(two rotational degrees of freedom) and 1 vibrational degree of freedom

The average energy of the molecule at temperature T is given by

By law of equipartition of energy we have

Thus the total energy per mole of the non-rigid diatomic gas is given by

Where R = universal gas constant and T = absolute temperature of the gasQ4. State the characteristics of the spectrum of black body radiations in terms of wavelength.

• The emissive power of a perfectly black body increases with an increase in its temperature for every wavelength.
• Each curve has characteristic form and each of them has a maxima i.e. maximum emissive power corresponding to a certain wavelength.
• The position of maxima shifts towards ultraviolet region (shorter wavelength) with the increase in the temperature.
• λm T = Constant (Wien’s displacement law)
• The area under each curve gives the total radiant power per unit area of the black body at that temperature and it is directly proportional to T4 (Verification of Stefan’s law)

Q4. State and explain Stefan’s law of radiation.

• The heat energy radiated per unit time per unit area of a perfectly black body is directly proportional to the fourth power of its absolute temperature.
• Explanation: Let Eb, the heat radiated per unit time per unit area of a perfectly black body whose absolute temperature is T.

So by Stefan’s Law,

Eb ∝ T4

Eb   = σ  T4

where σ  is a constant known as Stefan’s constant.

Q5. State and explain Newton’s law of cooling.

• The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.
• Explanation: Let θ and θo, be the temperature of a body and its surroundings respectively. Let dQ / dt be the rate of loss of heat, So from Newton’s Law of Cooling,

where k is a constant.

Q6. Assuming the expression for pressure P exerted by ideal gas, prove that the kinetic energy per unit volume is(3/2)P.

From kinetic theory of gases, the pressure exerted by a gas is given by

Where      M = Total mass of the gas = Nm

V = Volume of the gas

r  = Density of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

c = r.m.s. velocity of gas molecules.

This is an expression for the kinetic energy of gas molecules per unit volume of the gas.

Q7. Show that the r.m.s. velocity of gas molecules is directly proportional to the square root of its absolute temperature

• From kinetic theory of gases, the pressure exerted by a gas is given by

Where      M = Total mass of the gas = Nm

V = Volume of the gas

r  = Density of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

c = r.m.s. velocity of gas molecules.

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight.

Now R is universal gas constant, molecular mass M for a particular gas is constant.

Thus, the R.M.S. velocity of a gas is directly proportional to the square root of the absolute temperature of the gas.

Q8. A refrigerator is the reverse of a heat engine. Explain.

• In a heat engine, heat is transferred from a higher temperature level called source to a lower temperature level called sink. Work is obtained during this process.
•  In refrigerator the heat is transferred from the lower temperature level to higher temperature by applying external work to maintain the temperature below atmospheric temperature. Thus a refrigerator is a reversed heat engine.

Q9. What is the athermanous substance? Write two examples.

• The substances which cannot transmit the radiant heat incident upon their surfaces are called a diathermanous (athermanous) substances.
• e.g. wood, iron copper etc.

Q10. What is diathermanous substance? Write two examples.

• The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances.
• e.g. glass, quartz, gases

Q11. On what factors does the emissive power of a body depend?

• Emissive power of a body depends on a) the temperature of a body, b) the nature of the body, c) the surface area of the body, and d) the nature of the surroundings.

Q12. Why is it cold at the top of the mountain compared to sea level?

• As air rises, the pressure decreases. There is a lower pressure at higher altitudes than that at sea level.
• The lower pressure at the top of the mountain causes the temperature to be colder on top of a mountain than at sea level.

Q13. What is basic law followed by equipartition of energy?

• The Law of Equipartition of Energy defines the allocation of energy to each motion of the atom (translational, rotational and vibrational).
• In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to

Q14. Equal masses of monoatomic and diatomic gases are supplied heat at the same temperature, pressure, and volume. If the same amount of heat is supplied to both the gases, which of them will undergo a greater temperature rise? Why?

• For monoatomic gas, the temperature rise will be greater because molecules of monoatomic gas possess only a translational degree of freedom.
• whereas molecules of diatomic gas possess translation, rotation and vibrational (at higher temperature), so temperature size for diatomic gases is lower.

Q15. A gas is contained in a closed vessel. How pressure due to the gas will be affected if force between the molecules disappear suddenly

• As force of attraction between molecules disappears, then the randomness of motion of gas particles increases further. Thus the molecules of gas will hit the wall with more speeds, hence, the rate of change of momentum will increase.
• Now the increase in the rate of change of momentum of molecules means more force on the area of the wall of the container. Now pressure p = F/A will increase. Hence the pressure exerted by the gas on the walls of container increases.

### S.A. II (3 Marks)

Q1. What are degrees of freedom? Explain degrees of freedom of (a) a monoatomic and (b) a diatomic molecule.

• A molecule free to move in space needs three coordinates to specify its location.
• If a molecule is constrained to move along a line it requires one co-ordinate to locate it. Thus it has one degree of freedom for motion in a line. If a molecule is constrained to move in a plane it requires two coordinates to locate it. Thus it has two degrees of freedom for motion in a plane. If a molecule is free to move in a space it requires three coordinates to locate it. Thus molecules of monoatomic gases have three degrees of freedom for motion in a space.
• The molecules of monoatomic gases have only translational motion, hence has three translational degrees of freedom.
• The molecules of diatomic gases have translational motion ( three translational degrees of freedom) as well as rotational motion(two rotational degrees of freedom). Thus diatomic molecules have total 5 degrees of freedom.

Q2. Explain Maxwell’s distribution of molecular speeds with the suitable graph.

• For a given mass of a gas, the velocities of all molecules are not the same, even when bulk parameters like pressure, volume and temperature are fixed. Collisions change the direction and speed of the molecules, but in a state of equilibrium, the distribution of speed is constant.
• It is observed that the molecular speed distribution gives the number of molecules dN(v) between speeds v and v + dv, which is proportional to dv (difference in velocities) is called Maxwell distribution.
• Following graph shows the distribution of velocity at different temperatures

• The fraction of the molecules with speed v and v + dv is equal to the area of the strip shown.

Q3.  What is the coefficient of performance of a refrigerator? Derive an expression for it.

• The coefficient of performance (α) of a refrigerator is defined as the ratio of the heat extracted (Q2)from the closed reservoir to work done (W) on the system. Mathematically

α = Q2/W ………. (1)

• The efficiency (η) cannot exceed 1 but α can be greater than 1. By the law of conservation of energy

Q1 = W + Q2

∴  W = Q1 – Q2

Substituting in equation (1) we get

α = Q2/(Q1 – Q2)

This is an expression for coefficient of performance (COP) of a refrigerator.

Q4. State the law of equipartition of energy. Obtain an expression for energy per molecule and energy per mole of a monoatomic gas.

• In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to

• The molecules of monoatomic gases have only translational motion, hence has three translational degrees of freedom. The average energy of the molecule at temperature T is given by

By law of equipartition of energy we have

Thus the energy per mole of the gas is given by

Where R = universal gas constant and T = absolute temperature of the gas

Q5.  State the law of equipartition of energy. Obtain an expression for energy per molecule and energy per mole of a rigid diatomic gas.

• In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to

• The molecules of rigid diatomic gases have translational motion ( three translational degrees of freedom) as well as rotational motion(two rotational degree of freedom)

The average energy of the molecule at temperature T is given by

By law of equipartition of energy we have

Thus the energy per mole of the gas is given by

Where R = universal gas constant and T = absolute temperature of the gas

Q6. State the law of equipartition of energy. Obtain an expression for energy per molecule and energy per mole of a non-rigid diatomic gas.

• In equilibrium, the total energy is equally distributed in all possible energy modes, with vibrational mode has KBT of energy and each other mode having an average energy equal to

• The molecules of non-rigid diatomic gases have translational motion ( three translational degrees of freedom) as well as rotational motion(two rotational degrees of freedom) and 1 vibrational degree of freedom

The average energy of the molecule at temperature T is given by

By law of equipartition of energy we have

Thus the energy per mole of the non-rigid diatomic gas is given by

Where R = universal gas constant and T = absolute temperature of the gas

Q7. Explain Ferry’s perfectly black body with the help of a neat labelled diagram.

• A body which absorbs all the radiant heat incident upon it is called a perfectly black body.

Construction:

• A perfectly black body can be artificially constructed by taking a double-walled, hollow metal sphere Having a small hole.
• The inner surface of the sphere is coated with lamp black and it has a conical projection on the opposite side of the hole.

Working:

• The radiation entering the sphere through this hole suffers multiple reflections.
• During every reflection, about 98% of the incident radiant heat is absorbed by the sphere. Therefore the radiation is completely absorbed by the sphere within a few reflections.
• In this way, the sphere acts as a perfectly black body whose effective area is equal to the area of the hole.

Q8. Derive theoretically the Kirchhoff’s law of radiation.

• The ratio of the emissive power to the coefficient of absorption is constant for all substances at a given temperature and is equal to the emissive power of a perfectly black body at that temperature.   OR At any given temperature, the emissivity (or coefficient of emission) of a body is equal to the coefficient of absorption
• Explanation: If ‘E’ is the emissive power of a substance and ‘a’ is its coefficient of absorption then by Kirchoff’s law

i.e a = e

Theoretical Proof of Kirchoff’s Law of Radiation:

• Let us consider two bodies A and B suspended in a constant temperature enclosure. B is a perfectly black body. After sometime both A and B will attain the same temperature as that of enclosure. By Prevost heat exchange theory In this state also every body will emit and absorb thermal radiations.
• Let E be the emissive power of A and  ‘a’ be its coefficient of absorption. Let Eb be the emissive power of B. Let Q be the radiant heat incident per unit time per unit area of each body.

Heat absorbed per unit time per unit area of   A =   a Q

Heat  emitted per unit time per unit area  of A  =  E.

As the temperature remains constant so the heat emitted will be equal to the heat absorbed

∴   E    =  a Q      …………(1)

Perfectly black body B will absorb all the radiant heat incident on it.

Heat absorbed per unit time per unit area of  B  =   Q.

Heat emitted per unit time per unit area of B = Eb

As the temperature of B remains constant so in case of B also heat emitted is equal to heat absorbed.

Eb  = Q   …………………..(2)

Dividing equation (1) by (2) we get,

Thus, the coefficient of emission is equal to the coefficient of absorption.  This proves Kirchoff’s law.

Q9.  Derive an expression for pressure exerted by an ideal gas on any one wall of a container.

• Let us consider a gas enclosed in a cube whose each edge is of length ‘l’. Let A be the area of each face of the cube. So A  = . Let V be the volume of the cube (or the gas). So  V = . Let ‘m’ be the mass of each molecule of the gas and ‘N’ be the total number of molecules of the gas and ‘M’ be the total mass of the gas. So M = mN.
• Suppose that the three intersecting edges of the cube are along the rectangular co-ordinate axes X, Y and Z with the origin O at one of the corners of the cube.  By the kinetic theory of gases, we know that molecules of a gas are in a state of random motion so it can be imagined that on an average N/3 molecule are constantly moving parallel to each edge of the cube i.e. along the co-ordinate axes. Let velocities of N/3 molecules moving parallel to X-axis be C1, C2, C3, …. , CN/3 respectively.
• Consider a molecule moving with the velocity  C1in the positive direction of X-axis.

Initial momentum of the molecule,   p1 =  mC1,

It will collide normally with the wall ABCD and as the collision is perfectly elastic  rebounds with the same velocity.

∴ momentum  of molecule  after  collision, p2 =  – mC1

Change in momentum of molecule  due to one collision with ABCD = Δp =  p2  –   p1

Δp = – mC1  – mC1     =  – 2 – mC1

Before the next collision with the wall ABCD the molecule will travel a distance 2l with velocity C1.

So time interval between two successive collisions of the molecule with ABCD  = 2l /C1

Number of collisions of the molecule per unit with wall ABCD  =  C1/ 2l

Change   in   momentum    of the molecule  per   unit   time = –  2mC1  ×  C1/ 2l   = –  mC1² /l

But by Newton’s second law we know that the rate of change of momentum is equal to impressed force.

So force exerted on the molecule by wall ABCD  =-  mC1² /l

From    Newton’s   third law of motion, Action reaction are equal and opposite

So force exerted on the molecule  by wall ABCD   =  mC1² /l

Thus every molecule will exert a force on the wall ABCD.

So total force exerted on the wall ABCD due to molecules moving in positive X-axis direction is given by

Pressure    exerted    on    the   wall   ABCD     is given by

The pressure exerted on each wall will be the same and i.e. equal to the pressure of the gas.

Let P be the pressure of the gas.

Also by definition of r.m.s. velocity

This is an expression for pressure exerted by a gas on the walls of the container

 Maharashtra State Board > Science >  Kinetic Theory of Gases > You are Here